Question

At a point $A$ , the angle of elevation of a tower is found to be such that its tangent is $\frac{5}{12}$ , on walking $240\,m$ nearer to the tower, the tangent of the angle of elevation is found to be $\frac{3}{4}$ . What is the height of the tower ( in $m$ )?

Answer 1

Let $CO$ is the tower and $AB=240\,m$

$\tan\theta =\frac{C O}{A O}=\frac{C O}{240 + B O}=\frac{5}{12}$
$CO=\left(240 + B O\right)\frac{5}{12}...\left(1\right)$
$\tan\phi=\frac{C O}{O B}=\frac{3}{4}$
or $CO=\frac{3}{4}OB...\left(2\right)$
From Equations $\left(1\right)\&\left(2\right)$ ,
$\frac{3}{4}OB=\left(240 + B O\right)\frac{5}{12}$
or $9OB=1200+5BO$ ,
or $OB=300\,m$
$\therefore $ $CO=\frac{3}{4}\times 300=225\,m$
$\therefore $ Hence, height of the tower is $225\,m$ .